Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{35y - 28}{y} \times \dfrac{4y}{5(5y - 4)} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ (35y - 28) \times 4y } { y \times 5(5y - 4) } $ $ r = \dfrac {4y \times 7(5y - 4)} {y \times 5(5y - 4)} $ $ r = \dfrac{28y(5y - 4)}{5y(5y - 4)} $ We can cancel the $5y - 4$ so long as $5y - 4 \neq 0$ Therefore $y \neq \dfrac{4}{5}$ $r = \dfrac{28y \cancel{(5y - 4})}{5y \cancel{(5y - 4)}} = \dfrac{28y}{5y} = \dfrac{28}{5} $